Question

A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction between the hockey puck and the metal ramp are μs = 0.40 and μk = 0.30, respectively. The puck's initial speed is 99 m/s. What vertical height does the puck reach above its starting point?

Answer 1

the vertical height reach by the puck is 329.06m

Step-by-step explanation:

In all the process, the only non-conservative force presented in the problem is the frictional force. Therefore, applying the Mechanical energy conservation:

`\Delta E_(M) =W_(ncf)`

with:

`E_(Mi)=(1)/(2) mv_(i) ^(2)\\E_(Mf)=mgH`

`W_(ncf)=\int\limits^L_0 {\vec{F_(roz)} } \,\cdot \vec{dx}=-|F_(roz)|L=-(H|F_(roz)|)/(sin(\alpha))`

From the dynamic analysis:

`F_(roz)=\mu_(k)N=\mu_(k)cos(\alpha)mg`

Therefore:

`E_(Mf)-E_(Mi)=W_(ncf)`

`mgH-(1)/(2) mv_(i) ^(2)=-(Hmg\mu_(k))/(tan(\alpha))\\H-(v_(i) ^(2))/(2g)=-(H\mu_(k))/(tan(\alpha))\\H(1+(\mu_(k))/(tan(\alpha)))=(v_(i) ^(2))/(2g)\\H=(v_(i) ^(2))/(2g(1+(\mu_(k))/(tan(\alpha))))\\H=329.06m`

Answer 2

H=1020.12m

Step-by-step explanation:

From a balance of energy:

`(m*Vo^2)/(2) -mg*H=-Ff*d` where H is the height it reached, d is the distance it traveled along the ramp and Ff = μk*N.

The relation between H and d is given by:

H = d*sin(30) Replace this into our previous equation:

`(m*Vo^2)/(2) -mg*d*sin(30)=-\mu_k*N*d`

From a sum of forces:

N -mg*cos(30) = 0 => N = mg*cos(30) Replacing this:

`(m*Vo^2)/(2) -mg*d*sin(30)=-\mu_k*mg*cos(30)*d` Now we can solve for d:

d = 2040.23m

Thus H = 1020.12m